JEE Advanced 2013 Paper 1Question 1
Let complex numbers $\alpha$ and $\frac{1}{\alpha}$ lie on circles $(x-x_0)^2+(y-y_0)^2=r^2$ and $(x-x_0)^2+(y-y_0)^2=4r^2$, respectively. If $z_0=x_0+iy_0$ satisfies the equation $2|z_0|^2=r^2+2$, then $|\alpha|=$
Complex Numbers
Solution
Let $\rho=|\alpha|$. From $|\alpha-z_0|^2=r^2$: $2\operatorname{Re}(\alpha\bar z_0)=\rho^2+|z_0|^2-r^2$. The point $1/\alpha$ (taken with the same direction as $\alpha$, i.e. $\alpha/\rho^2$) gives $\frac{1}{\rho^2}-\frac{2\operatorname{Re}(\alpha\bar z_0)}{\rho^2}+|z_0|^2=4r^2$. Substituting and using $|z_0|^2=\frac{r^2+2}{2}$ gives $r^2(1+\rho^2)/2=4r^2\rho^2$, so $7\rho^2=1$ and $|\alpha|=\frac1{\sqrt7}$.